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- 1. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 1
The Normal
Distribution
Chapter 6
- 2. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 2
Objectives
In this chapter, you learn:
To compute probabilities from the normal distribution
How to use the normal distribution to solve business
problems
To use the normal probability plot to determine whether
a set of data is approximately normally distributed
- 3. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 3
Continuous Probability Distributions
A continuous variable is a variable that can
assume any value on a continuum (can assume
an uncountable number of values)
thickness of an item
time required to complete a task
temperature of a solution
height, in inches
These can potentially take on any value
depending only on the ability to precisely and
accurately measure
- 4. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 4
Bell Shaped
Symmetrical
Mean, Median and Mode
are Equal
Location is determined by the
mean, μ
Spread is determined by the
standard deviation, σ
The random variable has an
infinite theoretical range:
+ to
Mean
= Median
= Mode
X
f(X)
μ
σ
The Normal Distribution
- 5. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 5
The Normal Distribution
Density Function
2
μ)
(X
2
1
e
2π
1
f(X)
The formula for the normal probability density function is
Where e = the mathematical constant approximated by 2.71828
π = the mathematical constant approximated by 3.14159
μ = the population mean
σ = the population standard deviation
X = any value of the continuous variable
- 6. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 6
A
B
C
A and B have the same mean but different standard deviations.
B and C have different means and different standard deviations.
By varying the parameters μ and σ, we
obtain different normal distributions
- 7. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 7
The Normal Distribution Shape
X
f(X)
μ
σ
Changing μ shifts the
distribution left or right.
Changing σ increases
or decreases the
spread.
- 8. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 8
The Standardized Normal
Any normal distribution (with any mean and
standard deviation combination) can be
transformed into the standardized normal
distribution (Z)
To compute normal probabilities need to
transform X units into Z units
The standardized normal distribution (Z) has a
mean of 0 and a standard deviation of 1
- 9. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 9
Translation to the Standardized
Normal Distribution
Translate from X to the standardized normal
(the “Z” distribution) by subtracting the mean
of X and dividing by its standard deviation:
σ
μ
X
Z
The Z distribution always has mean = 0 and
standard deviation = 1
- 10. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 10
The Standardized Normal
Probability Density Function
The formula for the standardized normal
probability density function is
Where e = the mathematical constant approximated by 2.71828
π = the mathematical constant approximated by 3.14159
Z = any value of the standardized normal distribution
2
(1/2)Z
e
2π
1
f(Z)
- 11. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 11
The Standardized
Normal Distribution
Also known as the “Z” distribution
Mean is 0
Standard Deviation is 1
Z
f(Z)
0
1
Values above the mean have positive Z-values.
Values below the mean have negative Z-values.
- 12. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 12
Example
If X is distributed normally with mean of $100
and standard deviation of $50, the Z value
for X = $200 is
This says that X = $200 is two standard
deviations (2 increments of $50 units) above
the mean of $100.
2.0
$50
100
$
$200
σ
μ
X
Z
- 13. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 13
Comparing X and Z units
Note that the shape of the distribution is the same,
only the scale has changed. We can express the
problem in the original units (X in dollars) or in
standardized units (Z)
Z
$100
2.0
0
$200 $X(μ = $100, σ = $50)
(μ = 0, σ = 1)
- 14. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 14
Probability is measured by the area under
the curve
a b X
f(X) P a X b
( )
≤
≤
P a X b
( )
<
<
=
(Note that the
probability of any
individual value is zero)
Finding Normal Probabilities
- 15. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 15
Probability as
Area Under the Curve
The total area under the curve is 1.0, and the curve is
symmetric, so half is above the mean, half is below
f(X)
X
μ
0.5
0.5
1.0
)
X
P(
0.5
)
X
P(μ
0.5
μ)
X
P(
- 16. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 16
The Standardized Normal Table
The Cumulative Standardized Normal table in
the textbook (Appendix table E.2) gives the
probability less than a desired value of Z (i.e.,
from negative infinity to Z)
Z
0 2.00
0.9772
Example:
P(Z < 2.00) = 0.9772
- 17. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 17
The Standardized Normal Table
The value within the
table gives the
probability from Z =
up to the desired Z
value
.9772
2.0
P(Z < 2.00) = 0.9772
The row shows
the value of Z
to the first
decimal point
The column gives the value of
Z to the second decimal point
2.0
.
.
.
(continued)
Z 0.00 0.01 0.02 …
0.0
0.1
- 18. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 18
General Procedure for Finding
Normal Probabilities
Draw the normal curve for the problem in
terms of X
Translate X-values to Z-values
Use the Standardized Normal Table
To find P(a < X < b) when X is
distributed normally:
- 19. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 19
Finding Normal Probabilities
Let X represent the time it takes (in seconds) to
download an image file from the internet.
Suppose X is normal with a mean of18.0
seconds and a standard deviation of 5.0
seconds. Find P(X < 18.6)
18.6
X
18.0
- 20. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 20
Finding Normal Probabilities
Let X represent the time it takes, in seconds to download an image file from
the internet.
Suppose X is normal with a mean of 18.0 seconds and a standard deviation
of 5.0 seconds. Find P(X < 18.6)
Z
0.12
0
X
18.6
18
μ = 18
σ = 5
μ = 0
σ = 1
(continued)
0.12
5.0
8.0
1
18.6
σ
μ
X
Z
P(X < 18.6) P(Z < 0.12)
- 21. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 21
Z
0.12
0.5478
Standardized Normal Probability
Table (Portion)
0.00
= P(Z < 0.12)
P(X < 18.6)
Z .00 .01
0.0 .5000 .5040 .5080
.5398 .5438
0.2 .5793 .5832 .5871
0.3 .6179 .6217 .6255
.02
0.1 .5478
Solution: Finding P(Z < 0.12)
- 22. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 22
Finding Normal
Upper Tail Probabilities
Suppose X is normal with mean 18.0
and standard deviation 5.0.
Now Find P(X > 18.6)
X
18.6
18.0
- 23. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 23
Finding Normal
Upper Tail Probabilities
Now Find P(X > 18.6)…
(continued)
Z
0.12
0
Z
0.12
0.5478
0
1.000 1.0 - 0.5478
= 0.4522
P(X > 18.6) = P(Z > 0.12) = 1.0 - P(Z ≤ 0.12)
= 1.0 - 0.5478 = 0.4522
- 24. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 24
Finding a Normal Probability
Between Two Values
Suppose X is normal with mean 18.0 and
standard deviation 5.0. Find P(18 < X < 18.6)
P(18 < X < 18.6)
= P(0 < Z < 0.12)
Z
0.12
0
X
18.6
18
0
5
8
1
18
σ
μ
X
Z
0.12
5
8
1
18.6
σ
μ
X
Z
Calculate Z-values:
- 25. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 25
Z
0.12
0.0478
0.00
= P(0 < Z < 0.12)
P(18 < X < 18.6)
= P(Z < 0.12) – P(Z ≤ 0)
= 0.5478 - 0.5000 = 0.0478
0.5000
Z .00 .01
0.0 .5000 .5040 .5080
.5398 .5438
0.2 .5793 .5832 .5871
0.3 .6179 .6217 .6255
.02
0.1 .5478
Standardized Normal Probability
Table (Portion)
Solution: Finding P(0 < Z < 0.12)
- 26. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 26
Probabilities in the Lower Tail
Suppose X is normal with mean 18.0
and standard deviation 5.0.
Now Find P(17.4 < X < 18)
X
17.4
18.0
- 27. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 27
Probabilities in the Lower Tail
Now Find P(17.4 < X < 18)…
X
17.4 18.0
P(17.4 < X < 18)
= P(-0.12 < Z < 0)
= P(Z < 0) – P(Z ≤ -0.12)
= 0.5000 - 0.4522 = 0.0478
(continued)
0.0478
0.4522
Z
-0.12 0
The Normal distribution is
symmetric, so this probability
is the same as P(0 < Z < 0.12)
- 28. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 28
Empirical Rule
μ ± 1σ encloses about
68.26% of X’s
f(X)
X
μ μ+1σ
μ-1σ
What can we say about the distribution of values
around the mean? For any normal distribution:
σ
σ
68.26%
- 29. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 29
The Empirical Rule
μ ± 2σ covers about 95.44% of X’s
μ ± 3σ covers about 99.73% of X’s
x
μ
2σ 2σ
x
μ
3σ 3σ
95.44% 99.73%
(continued)
- 30. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 30
Given a Normal Probability
Find the X Value
Steps to find the X value for a known
probability:
1. Find the Z value for the known probability
2. Convert to X units using the formula:
Zσ
μ
X
- 31. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 31
Finding the X value for a Known
Probability
Example:
Let X represent the time it takes (in seconds) to
download an image file from the internet.
Suppose X is normal with mean 18.0 and standard
deviation 5.0
Find X such that 20% of download times are less than
X.
X
? 18.0
0.2000
Z
? 0
(continued)
- 32. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 32
Find the Z value for
20% in the Lower Tail
20% area in the lower
tail is consistent with a
Z value of -0.84
Z .03
-0.9 .1762 .1736
.2033
-0.7 .2327 .2296
.04
-0.8 .2005
Standardized Normal Probability
Table (Portion)
.05
.1711
.1977
.2266
…
…
…
…
X
? 18.0
0.2000
Z
-0.84 0
1. Find the Z value for the known probability
- 33. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 33
Finding the X value
2. Convert to X units using the formula:
8
.
13
0
.
5
)
84
.
0
(
0
.
18
Zσ
μ
X
So 20% of the values from a distribution
with mean 18.0 and standard deviation
5.0 are less than 13.80
- 34. Copyright © 2016 Pearson Education, Ltd. Chapter 6, Slide 34
Chapter Summary
In this chapter we discussed:
Computing probabilities from the normal distribution
Using the normal distribution to solve business
problems
Using the normal probability plot to determine whether a
set of data is approximately normally distributed